My input matrix does not contain zeros but i get zeros when I print the matrix.The problem description is given below

Matrix Layer Rotation(hacker rank problem)

You are given a 2D matrix of dimension m X n and a positive integer r. You have to rotate the matrix times and print the resultant matrix. Rotation should be in the anti-clockwise direction.

The rotation of a 4 X 5 matrix is represented by the following figure. Note that in one rotation, you have to shift elements by one step only.

matrix-rotation

It is guaranteed that the minimum of m and n will be even.

As an example rotate the Start matrix by 2:

Start         First           Second
1 2 3 4        2  3  4  5      3  4  5  6
12 1 2 5  ->   1  2  3  6 ->   2  3  4  7
11 4 3 6      12  1  4  7       1  2  1  8
10 9 8 7      11 10  9  8     12 11 10  9

#include<stdio.h>
#include<stdlib.h>
int main()
{
int d=0,m,n,r;
long i,j;
scanf("%d %d %d",&m,&n,&r);
long **a;
a=(long**)malloc(m*sizeof(long*));
for(i=0;i<m;i++)
{
a[i]=(long*)malloc(n*sizeof(long));
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%ld",&a[i][j]);
}
}
while(d<r)
{
int lr=m-1,lc=n-1,ro=0,c=0;
long **b;
b=(long**)malloc(m*sizeof(long*));
for(i=0;i<m;i++)
{
b[i]=(long*)malloc(n*sizeof(long));
}
while(ro<lr && c<lc)
{
if(c<lc)
{
for(i=lc;i>=c;i--)
{
if(i==lc)
{
b[ro][i]=a[ro+1][lc];
}
else
{
b[ro][i]=a[ro][i+1];
}
}
ro++;
}
for(i=ro;i<=lr;i++)
{
b[i][c]=a[i-1][c];
}
c++;
for(i=c;i<=lc;i++)
{
b[lr][i]=a[lr][i-1];
}
lr--;
if(ro<lr)
{
for(i=lr;i>=r;i--)
{
b[i][lc]=a[i+1][lc];
}
lc--;
}
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
a[i][j]=b[i][j];
}
}
d++;
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
printf("%ld ",a[i][j]);
}
printf("\n");
}
return 0;
}

Compiler Message
Wrong Answer
Input (stdin)
5 4 7
1 2 3 4
7 8 9 10
13 14 15 16
19 20 21 22
25 26 27 28
Your Output (stdout)
0 0 0 0 
0 0 0 0 
0 0 0 0    (0 is the problem)
10 0 0 0 
4 3 2 1 
Expected Output
28 27 26 25
22 9 15 19
16 8 21 13
10 14 20 7
4 3 2 1

test case 2
Wrong Answer
Input (stdin)
5 4 7
1 2 3 4
7 8 9 10
13 14 15 16
19 20 21 22
25 26 27 28
Your Output (stdout)
28 27 26 25 
22 0 0 19 
16 0 0 13 
10 0 0 7 
4 3 2 1 
Expected Output
28 27 26 25
22 9 15 19
16 8 21 13
10 14 20 7
4 3 2 1

The code within your while(d<r) loop should accomplish a single position rotation of a each time it is executed. while(d<r) { ... d++; } then causes the body of the loop to be executed r times.

The issue that you're having likely arises on this loop:

for(i=lr;i>=r;i--) {
b[i][lc]=a[i+1][lc];
}

with i>=r. Nothing in the body of the loop should be a function of either d or r. A single position rotation should not depend on how many single rotations will be performed (or have already been performed). Clear?

--- Adding my own implementation of what I think you're asking for ---

The following seems to do what you need. In any case, I've commented it to make the purpose of each operation more clear. Feel free to use it if it works for you.

#include <stdio.h>
#include <stdlib.h>
int main()
{
int m, n, repetitions;
long i, j;
scanf("%d %d %d", &m, &n, &repetitions);
long **a;
a = (long **)malloc(m * sizeof(long *));
for (i = 0; i < m; i++) {
a[i] = (long *)malloc(n * sizeof(long));
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
scanf("%ld", &a[i][j]);
}
}
long **b;
b = (long **)malloc(m * sizeof(long *));
for (i = 0; i < m; i++) {
b[i] = (long *)malloc(n * sizeof(long));
}
while (repetitions--) {
// Upper-left, lower-row, lower-column
int ul = 0, lr = m - 1, lc = n - 1;
// Perform a single position rotation, from the outer ring working in. Stop
// when either the width or height of the ring is 1.
while (ul < lr && ul < lc) {
// Shift left-edge downwards
for (int rr = ul; rr < lr; ++rr) b[rr + 1][ul] = a[rr][ul];
// Shift bottom-edge rightwards
for (int cc = ul; cc < lc; ++cc) b[lr][cc + 1] = a[lr][cc];
// Shift right-edge upwards
for (int rr = lr; rr > ul; --rr) b[rr - 1][lc] = a[rr][lc];
// Shift top-edge leftwards
for (int cc = lc; cc > ul; --cc) b[ul][cc - 1] = a[ul][cc];
// Shift upper-left corner of ring to the right and down by 1.
++ul;
// Shift bottom-edge of ring up by 1.
--lr;
// Shift right-edge of ring to the left by 1.
--lc;
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
a[i][j] = b[i][j];
}
}
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%ld ", a[i][j]);
}
printf("\n");
}
return 0;
}